# algebraic closure is the intersection of all elementary sub-models of the monster

This is a question from an exercise in model theory. Let T be a complete theory, $\mathfrak{C}$ monster model of T (a $\kappa$ saturated model of cardinality $\kappa$ for some large $\kappa$) and $A\subseteq|\mathfrak{C}|$. The algebraic closure of A is the union of all finite definable (over A) subsets of $|\mathfrak{C}|$. I need to prove that $$acl(A)=\bigcap \big\{ M\prec\mathfrak{C} | A\subseteq|M| \big\}$$ I was able to show the inclusion of acl(A) in any elementary submodel of $\mathfrak{C}$ that contains A but where not able to show the other inclusion.
There is a hint that we need to use another question in which I proved that the elements of acl(A) are exactly the elements that have finite number of conjugates over A ($c\in acl(A)$ iff the orbit of c under the operation of $Aut({\mathfrak{C}/A})$ is finite).
Suppose $c\in acl(A)$, then we need to find $M\prec\mathfrak{C}$ s.t. $A\subseteq|M|$ but $c\notin M$. I have tried to find a model that does not contain any of c's conjugates until I have found an example in which no such model exist.

## 1 Answer

Suppose $c\notin \text{acl}(A)$. Let $M \prec \mathfrak{C}$ be any small model with $A\subseteq M$. Can you find a realization $c'$ of $\text{tp}(c/A)$ which is not in $M$? (Use compactness and saturation of $\mathfrak{C}$.) Then moving $c'$ to $c$ by an automorphism fixing $A$ moves $M$ to a model $M'$ containing $A$, and $c\notin M'$.

• of course, it is obvious now :). thank you – sss89 Feb 3 '16 at 20:34